barcode generator for ssrs The sequence that has an N-point DFT equal to Y(k) = H(k)X(k) is in Software

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The sequence that has an N-point DFT equal to Y(k) = H(k)X(k) is
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CHAP. 61
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THE DFT
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Because we would like to express y(n) in terms of x(n) and h(n), let us substitute
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into the expression for y(n) as follows:
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Interchanging the order of the summations gives
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However, note that the term in brackets is equal to x((n - I)),. Therefore, it follows that
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n = 0, 1, . . . , N - I
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which is equivalent to
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as was to be shown.
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Let y(n) be Ihe linear convolution of the two finite-length sequences, h ( n )and r ( n ) ,of length N ,
and let y N ( n )be the N-point circular convolution
Derive the following relationship between y(n) and yN(n):
There are several ways toderive this relationship. One is to examine what happens when the DTFT of y(n) is sampled. Alternatively, this result may be derived from a systems point of view as follows. First, note that yN(n) is equal to one period of the linear. convolution of the finite-length sequence h(n) with the periodic sequencei(n):
If we let
then the periodic sequence .f(n) is formed by linearly convolving x ( n ) with pN(n):
Therefore, the N-point circular convolution may be written as
THE DFT which is illustrated in the figure below.
[CHAP. 6
Because the first three systems are linear and shift-invariant. the order of these systems may be interchanged as illustrated in the following figure:
However, note that the output of the second filter, y(n), is the linear convolution of h(n) with x(n). This sequence is then convolved with pN(n), which gives the periodic sequence
This sequence is then multiplied by the rectangular window R N ( n ) ,
which is the relationship that was to be established.
How may we compute the N-point DFT of two real-valued sequences, xl ( n ) and x2(n),using one N -point DFT
The DFTs of two real-valued sequences may be fbund from one N -point DFT as follows. First, we form the N-point complex sequence
After finding the N-point DFTofx(n), we extract X , ( k )and X2(k)from X(k) by exploiting the symmetry properties of the DFT. Specifically, recall that the DFT of a real-valued sequence is conjugate symmetric,
and the DFT of an imaginary sequence is conjugate antisymmetric,
Therefore, because X(k) = Xl(k) with X I(k) the DFT of a real-valued sequence, then
+ X (k)
which is the conjugate symmetric part of X(k). Similarly, because Xz(k) is the DFT of an imaginary sequence, Xz(k) = i [ ~ ( k- X*((N - k ) ) ~ l ) which is the conjugate antisymmetric part of X(k).
CHAP. 61
THE DFT
Let x l ( n ) and x z ( n ) be N-point sequences with N-point DFTs X l ( k ) and X z ( k ) , respectively. Find an expression for the N-point DFT of the product x ( n ) = xl(n).rz(n)in terms of X 1 ( k ) and X 2 ( k ) .
Just as we have seen with the DTFT, there is a duality in the DFT properties. We have seen, for example, that multiplying a sequence by a complex exponential results in a circular shift of the DFT coefficients. Similarly, multiplying the DFT coefficients by a complex exponential results in a circular shift of the sequence. Therefore, because multiplying DFTcoefficients corresponds to a circular convolution of the sequences, we expect that the multiplication of two sequences would result in the circular convolution of their DFTs. To establish this property, we begin by noting that
Because we would like to express X ( k ) in terms of X I ( k )and X 2 ( k ) ,we substitute the following expression for x z ( n ) into Eq. (6.21):
The result is
Interchanging the order of the summations, this becomes
Recognizing that the second sum is X I( ( k -
we have
Therefore, X ( k ) is I /N times the circular convolution of X I ( k ) with X z ( k ) :
If x l ( n ) and q ( n ) are N-point sequences with N-point DFTs X I ( k ) and X (k), respectively, show that
This result is easy to derive if we use the properties of the DFT thar we already have. First, note that if X ( k ) is the N-point DFT of x ( n ) = x l ( n ) x ; ( n ) , then
Second, note that the DFT of x;(n) is X;((-k)),v,
Finally. recall that if x ( n ) = .ul(n)x;(n), the N-point DFT of x ( n ) is I/N times the circular convolution of X l ( k ) and X ; ( ( - k ) ) , ~(see Prob. 6.19):
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